LintCode 83. Single Number II (Medium)
LeetCode 137. Single Number II (Medium)
以下算法的复杂度都是:
时间复杂度: O(n)
空间复杂度: O(1)
解法1. 32个计数器
最简单的思路是用32个计数器, 满3复位0.
class Solution {
public:
int singleNumberII(vector<int> &A) {
int cnt[32] = {0};
int res = 0;
for (int i = 0; i < 32; ++i) {
for (int n : A) {
cnt[i] += (n >> i) & 1;
cnt[i] %= 3;
}
res |= cnt[i] << i;
}
return res;
}
};
解法2. 找规律
我的思路是, 解法1中的计数其实只需要两个bit就够了. 所有的首个bit记做res, 所有的第二个bit记做carry, 找规律:
如果A[i]的第k位是0, 则res和carry的第k位保持原样.
如果A[i]的第k位是1, 则:
| res | carry | res' | carry' |
|---|---|---|---|
| 0 | 0 | 1 | 0 |
| 1 | 0 | 0 | 1 |
| 0 | 1 | 0 | 0 |
此时(A[i][k]=1时)的规律就是:res'[k]=~res[k] & ~carry[k]carry'[k]=res[k] & ~carry[k]
class Solution {
public:
int singleNumberII(vector<int> &A) {
int res = 0, carry = 0;
for (int n : A) {
for (int i = 0; i < 32; ++i) {
int mask = (1 << i);
int bit = n & mask;
if (bit) {
int newRes = (~mask & res) + ((~res & mask) & (~carry & mask));
carry = (~mask & carry) + ((res & mask) & (~carry & mask));
res = newRes;
}
}
}
return res;
}
};
解法2.1. 解法2的简化
解法2中, 将A[i][k]=0和=1的情况合并, 可以得到:res'[k]=(A[i][k] & ~res[k] & ~carry[k]) | (~A[i][k] & res[k])carry'[k]=(A[i][k] & res[k] & ~carry[k]) | (~A[i][k] & carry[k])
这样的好处是可以32位一起算, 而不用一位一位地算:res'=(A[i] & ~res & ~carry) | (~A[i][k] & res)carry'=(A[i] & res & ~carry) | (~A[i][k] & carry)
class Solution {
public:
int singleNumberII(vector<int> &A) {
int res = 0, carry = 0;
for (int n : A) {
int newRes = (n & (~res & ~carry)) | (~n & res);
carry = (n & (res & ~carry)) | (~n & carry);
res = newRes;
}
return res;
}
};
解法3. one, two, three
Discuss中看到的解法, 自己实在想不出来. 用one, two和three三个int值作为bit flags.
循环对A[0]到A[n]进行考察, 当考察A[i]时:
记S[i]={A[0],...,A[i]},
若S[i]中所有数字的第k位bit的数目%3==1, 则one的第k位为1, 否则为0.
若S[i]中所有数字的第k位bit的数目%3==2, 则two的第k位为1, 否则为0.
而three是一个临时变量, 用于记录这一轮中, 哪些bit的数目恰巧是3的倍数.
two |= one & n;: 给two加上那些从1到2的数字.one ^= n;: 这句比较巧妙, 既删掉了会变成2的那些1, 又加上了新的1.three = one & two;: 1+2=3...one&= ~three: 从1中刨去那些成为3的1.two&= ~three: 从2中跑去那些成为3的2.
...如果你能解释得更清晰易懂, 欢迎留言!
class Solution {
public:
int singleNumberII(vector<int> &A) {
int one = 0, two = 0, three = 0;
for (int n : A) {
two |= one & n;
one ^= n;
three = one & two;
one &= ~three;
two &= ~three;
}
return one;
}
};
解法3.1. 解法3的变形
自己没想出解法4, 但是参照它的思路, 写了一个对自己比较直观的算法.
three = two & n;: 算出从2变成3的那些2.two = (two & ~three) | (one & n);: (two & ~three)是从2中刨去那些变为3的2, (one & n)是加上那些从1变成2的1.one = (one & ~n) | (n & ~three & ~two);: (one & ~n)是从1中刨去那些变成2的1, (n & ~three & ~two)是加上那些没给"2变3"或"1变2"用过的1.
class Solution {
public:
int singleNumberII(vector<int> &A) {
int one = 0, two = 0;
for (int n : A) {
int three = two & n;
two = (two & ~three) | (one & n);
one = (one & ~n) | (n & ~three & ~two);
}
return one;
}
};